"""
难度：简单
给定单向链表的头指针和一个要删除的节点的值，定义一个函数删除该节点。

返回删除后的链表的头节点。

注意：此题对比原题有改动

示例 1:
输入: head = [4,5,1,9], val = 5
输出: [4,1,9]
解释: 给定你链表中值为 5 的第二个节点，那么在调用了你的函数之后，该链表应变为 4 -> 1 -> 9.
示例 2:
输入: head = [4,5,1,9], val = 1
输出: [4,5,9]
解释: 给定你链表中值为 1 的第三个节点，那么在调用了你的函数之后，该链表应变为 4 -> 5 -> 9.
"""

# Definition for singly-linked list.
class ListNode:
    def __init__(self, x):
        self.val = x
        self.next = None
class Solution:
    def deleteNode(self, head: ListNode, val: int) -> ListNode:
        newHead = ListNode(0,head)
        newHead.next = head

        pre, curr = newHead,head
        while curr:
            if curr.val == val:
                pre.next = curr.next
            else:
                pre =curr
            curr = curr.next
        return newHead.next
        

